Methods of Proof 341inscribed in an 8×8 square decomposed into 64 unit squares. Because 3^2 + 32 > 42 , the
four unit squares at the corners lie ouside the circle. The interior of the circle is therefore
covered by 60 squares, which are our “holes.’’ The 61 points are the “pigeons,’’ and by
the pigeonhole principle two lie inside the same square. The distance between them does
not exceed the length of the diagonal, which is
√
- The problem is solved.
Figure 4845.Ifr=1, all lines pass through the center of the square. Ifr =1, a line that divides
the square into two quadrilaterals with the ratio of their areas equal torhas to pass
through the midpoint of one of the four segments described in Figure 49 (in that figure
the endpoints of the segments divide the sides of the square in the ratior). Since there
are four midpoints and nine lines, by the pigeonhole principle three of them have to pass
through the same point.
Figure 4946.Choose a face with maximal number of edges, and letnbe this number. The number
of edges of each of thenadjacent faces ranges between 3 andn, so by the pigeonhole
principle, two of these faces have the same number of edges.
(Moscow Mathematical Olympiad)